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Math Help - integral help

  1. #1
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    integral help

    Hello, I am to do the following integral two different wants. Now one way that I picked, which was the easiest was product to sum trig identities. I found the correct answer. Now, Im trying to use integration by parts, and the general idea, is integrate until you get back to your same integral, and then algebraically solve but I can't seem to get there. Here is the problem:



    Now, I get to a point(and please correct me if I'm wrong) where I'm at



    But from there.. clearly I can't do anything, if I subtract to the other side, that would defeat my purpose by getting rid of the integral. Maybe I'm doing it completely wrong? Help would be much appreciated!
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  2. #2
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    Quote Originally Posted by intervade View Post
    Hello, I am to do the following integral two different wants. Now one way that I picked, which was the easiest was product to sum trig identities. I found the correct answer. Now, Im trying to use integration by parts, and the general idea, is integrate until you get back to your same integral, and then algebraically solve but I can't seem to get there. Here is the problem:



    Now, I get to a point(and please correct me if I'm wrong) where I'm at



    But from there.. clearly I can't do anything, if I subtract to the other side, that would defeat my purpose by getting rid of the integral. Maybe I'm doing it completely wrong? Help would be much appreciated!
    Please post your working. You've probably dropped a minus somewhere.
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  3. #3
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    Alright, so for



    using



    u = sin(4x)
    dv = cos(5x)dx
    du = 4cos(4x)dx
    v = (1/5)sin(5x)



    again..
    u = sin(5x)
    dv = cos(4x)dx
    du = 5cos(5x)dx
    v = (1/4)sin(4x)

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  4. #4
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    Quote Originally Posted by intervade View Post
    Alright, so for



    using



    u = sin(4x)
    dv = cos(5x)dx
    du = 4cos(4x)dx
    v = (1/5)sin(5x)



    again..
    u = sin(5x)
    dv = cos(4x)dx
    du = 5cos(5x)dx
    v = (1/4)sin(4x)

    Add \frac{5}{4} \int \sin(4x) \cos(5x) to both sides
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  5. #5
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    Hmm, don't I need to distribute the -(4/5)?
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  6. #6
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    Quote Originally Posted by intervade View Post
    Hmm, don't I need to distribute the -(4/5)?
    Oops you would, I missed that set of brackets >.<
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